Let I = Int sin^3x dx.

I = Int sin^2x*sinx dx

We use Int (u(x)*v(x) dx = u(x)Intv(x)dx - Int{u'(x) int v(x))dx}dx.

I = (sin^2x) Intsinx dx - Int {(sin^2x)' Int sinxdx}dx

I = sin^2x(-cosx) - Int {2sinxcosx(-cosx)}dx

I = -sin^2xcosx +2Int (sinxcos^2x)dx

I = -sin^2xcosx +2Int(sinx(1-sin^2x)dx

I = -sin^2xcosx +2Int sinxdx -2I.

Add 2I to both sides:

3I = -sin^2cosx +2(-cosx)

I = - (1/3)cosx(2+sin^2x)_

We'll write the function as a product:

(sinx)^3 = (sinx)^2*sin x

We'll integrate both sides:

Int (sinx)^3dx = Int [(sinx)^2*sin x]dx

We'll write (sinx)^2 = 1 - (cosx)^2

Int [(sinx)^2*sin x]dx = Int [(1 - (cosx)^2)*sin x]dx

We'll remove the brackets:

Int [(1 - (cosx)^2)*sin x]dx = Int sin xdx - Int (cosx)^2*sin xdx

We'll solve Int (cosx)^2*sin xdx using substitution technique:

cos x = t

We'll differentiate both sides:

cos xdx = dt

We'll re-write the integral, changing the variable:

Int (cosx)^2*sin xdx = Int t^2dt

Int t^2dt = t^3/3 + C

Int (cosx)^2*sin xdx = (cos x)^3/3 + C

Int (sinx)^3dx = Int sin xdx - Int (cosx)^2*sin xdx

**Int (sinx)^3dx = -cos x - (cos x)^3/3 + C**